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Let $X$ be a Banach space and $X^*$ its dual. I would like to better understand the Mackey topology $\tau(X^*, X)$ on $X^*$.

The Mackey topology on $X$, $\tau(X,X^*)$ is defined as the topology of uniform convergence on $\sigma(X^*, X)$-compact convex sets of $X^*$. Swapping $X$ and $X^*$ in this definition would suggest the Mackey topology on $X^*$ is the topology of uniform convergence on $\sigma(X, X^*)$-compact convex sets of $X$. Is $\sigma(X, X^*)$ here the ordinary weak topology on $X$ induced by $X^*$ or is this abuse of notation?

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Just to clearify that Mackey topology on a normed space $X$ is not a universal/unique topology. It is always associated to a pairing $\left<X,Y,b\right>$ where $Y$ is another normed space and $b:X\times Y\to \mathbb C$ is a bilinear map. For the canonical pairing $X\times X^*\to \mathbb C$ defined as $x\times x^*\mapsto x^*(x)$ we have Mackey topology on $X$ defined as the topology of uniform convergence on $\sigma(X^*,X)$ compact convex subsets of $X^*$. On $X^*$ you can have two different types of canonical pairing $X^*\times X^{**}\to \mathbb C$ and $X^*\times X\to \mathbb C$, and you can obtain distinct Mackey toplogies on $X^*$ with respect to both these pairings. Now its your choice which Mackey topology you need to solve your problem. Following wikipedia articles are self contained and interesting to learn more about Mackey topology.

https://en.wikipedia.org/wiki/Mackey_topology

https://en.wikipedia.org/wiki/Polar_topology

Coming back to your question, $\sigma(X,X^*)$ is the usual weak topology on $X$ generated by $X$ and it makes perfect sense to have a Mackey topology on $X^*$ as the topology of uniform convergence on compact convex subsets in $X$. Just view the elements of $X$ as the element of $X^{**}$.

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  • $\begingroup$ Thank you for your answer. $\endgroup$
    – CBBAM
    Commented Jun 8, 2023 at 17:31

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